Question #190986

The coefficient of x in the Maclaurin's expansion of sinx sink is


Expert's answer

Given, f(x)=sinxsinkf(x)=sinx sink


so

f′(x)=cosxsink⇒f′(0)=cos0sink=sinkf′′(x)=−sinxsink⇒f′′(0)=0f′′′(x)=−cosxsink⇒f′′′(0)=−sinkf'(x)=cosx sink\Rightarrow f'(0)=cos0 sink=sink \\ f''(x)=-sin xsink\Rightarrow f''(0)=0 \\ f'''(x)=-cosxsink\Rightarrow f'''(0)=-sink


Maclaurins expansion is given by-


f(x)=f(0)+xf′(0)+x22!f′′(0)+x33!f′′′(0)...f(x)=0+x(sink)+x22!×0+x33!(−sink)f(x)=f(0)+xf'(0)+\dfrac{x^2}{2!}f''(0)+\dfrac{x^3}{3!} f'''(0)... \\[9pt] f(x)=0+x(sink)+\dfrac{x^2}{2!}\times 0+\dfrac{x^3}{3!} (-sink)


The coefficient of x in the expansion is sink.

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